1. Goodness of Fit Test for Proportions of Multinomial Population Chi-square distribution Hypotheses test/Goodness of fit test

2. Chi-square distribution 0 0 With 2 degrees of freedom of freedom With 2 degrees of freedom of freedom With 5 degrees of freedom of freedom With 5 degrees of freedom of freedom With 10 degrees of freedom of freedom With 10 degrees of freedom of freedom

3. Chi-Square Distribution For example, there is a.95 probability of obtaining a  2 (chi-square) value such that We will use the notation to denote the value for the chi-square distribution that provides an area of  to the right of the stated value.

4. Selected Values from the Chi-Square Distribution Table Our value For 9 d.f. and  =.975

5. 22 22 0 0.025 Area in Upper Tail =.975 2.700

6. Selected Values from the Chi-Square Distribution Table For 9 d.f. and  =.025 Our value

7. 22 22 0 0 19.023 Area in Upper Tail =.025 Area in Upper Tail =.025

8. Selected Values from the Chi-Square Distribution Table For 9 d.f. and  =.10 Our value

9. 22 22 0 0 14.684 Area in Upper Tail =.10 Area in Upper Tail =.10

10. For 9 d.f. and =16.919, =.05 For 8 d.f. and =3.49, =.90 For 6 d.f. and =16.812, =.01 For 10 d.f. and =18.9, = between.05 and.025

11. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f i, for each of the k categories. frequency, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the frequency, e i, in each category by multiplying the category probability by the sample size. category probability by the sample size. This is simply a hypothesis test to see if the hypothesized population proportions agree with the observed population proportions from our sample.

12. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 4. Compute the value of the test statistic. Note: The test statistic has a chi-square distribution with k – 1 df provided that the expected frequencies are 5 or more for all categories. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

13. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population where  is the significance level and there are k - 1 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if

14. Multinomial Distribution Goodness of Fit Test Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures Finger Lakes Homes manufactures four models of prefabricated homes, four models of prefabricated homes, a two-story colonial, a log cabin, a a two-story colonial, a log cabin, a split-level, and an A-frame. To help split-level, and an A-frame. To help in production planning, management in production planning, management would like to determine if previous would like to determine if previous customer purchases indicate that there customer purchases indicate that there is a preference in the style selected. is a preference in the style selected.

15. Split- A- Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15 The number of homes sold of each The number of homes sold of each model for 100 sales over the past two years is shown below. Multinomial Distribution Goodness of Fit Test Example: Finger Lakes Homes (A)

16. n Hypotheses Multinomial Distribution Goodness of Fit Test where: p C = population proportion that purchase a colonial p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame p A = population proportion that purchase an A-frame H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p C =.25, p L =.25, p S =.25, and p A =.25 p C =.25, p L =.25, p S =.25, and p A =.25

17. n Rejection Rule 22 22 7.815 Do Not Reject H 0 Reject H 0 Multinomial Distribution Goodness of Fit Test With  =.05 and k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3 degrees of freedom degrees of freedom if p -value 7.815. Reject H 0 if p -value 7.815.

18. Expected Frequencies Test Statistic Multinomial Distribution Goodness of Fit Test e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 = 1 + 1 + 4 + 4 = 10

19. Multinomial Distribution Goodness of Fit Test Conclusion Using the p-Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 10 is between 9.348 and 11.345, the Because  2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.025 and.01..025 and.01. Area in Upper Tail.10.05.025.01.005  2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

20. Conclusion Using the Critical Value Approach Multinomial Distribution Goodness of Fit Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that there is no home style preference.  2 = 10 > 7.815